从初等变换的角度理解
\begin{equation*}
E(j,k)^TAE(j,k), E(j(c))^TAE(j(c)), E(j,k(c))^TAE(j,k(c)).
\end{equation*}
解答.
注意到
\begin{equation*}
E(j,k)^T=E(j,k), E(j(c))^T=E(j(c)), E(j,k(c))^T=E(k,j(c)),
\end{equation*}
所以
\begin{equation*}
E(j,k)^TAE(j,k)=\begin{pmatrix}
a_{11} & \cdots & a_{1k} & \cdots & a_{1j} & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{k1} & \cdots & a_{kk} & \cdots & a_{kj} & \cdots & a_{kn}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{j1} & \cdots & a_{jk} & \cdots & a_{jj} & \cdots & a_{jn}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & a_{nk} & \cdots & a_{nj} & \cdots & a_{nn}
\end{pmatrix}
\end{equation*}
是对换\(A\)的第\(j\)行与第\(k\)行,再对换第\(j\)列与第\(k\)列所得;
\begin{equation*}
E(j(c))^TAE(j(c))=\begin{pmatrix}
a_{11} & \cdots & ca_{1j} & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots\\
ca_{j1} & \cdots & c^2a_{jj} & \cdots & ca_{jn}\\
\vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & ca_{nj} & \cdots & a_{nn}
\end{pmatrix}
\end{equation*}
是将\(A\)的第\(j\)行乘以非零常数\(c\),再将第\(j\)列乘以相同的常数\(c\)所得;
\begin{equation*}
E(j,k(c))^TAE(j,k(c))=\begin{pmatrix}
a_{11} & \cdots & a_{1j} & \cdots & a_{1k}+ca_{1j} & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{j1} & \cdots & a_{jj} & \cdots & a_{jk}+ca_{jj} & \cdots & a_{jn}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{k1}+ca_{j1} & \cdots & a_{kj}+ca_{jj} & \cdots & a_{kk}+ca_{jk}+c\left(a_{kj}+ca_{jj}\right) & \cdots & a_{kn}+ca_{jn}\\
\vdots & & \vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & a_{nj} & \cdots & a_{nk}+ca_{nj} & \cdots & a_{nn}
\end{pmatrix}
\end{equation*}
是将\(A\)的第\(j\)行乘以\(c\)加到第\(k\)行上,再将第\(j\)列乘以\(c\)加到第\(k\)列上所得。